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In the figure shown,initially the spring is in relaxed position. A bullet of mass m hits the block 'A' with horizontal speed v_(0) and gets embedded into it. Find the maximum compression in the spring. Stiffness of the spring is K. Neglect any friction and deformation in the spring during collision. |
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Answer» Solution :SINCE no external force is associated with the system `(M+M+m)` conservation of horizontal MOMENTUM of the system yields. `mV_(0)=(M+m)v_(1)=(2M+m)v_(2)` Where `v_(1)=` velocity of (M+m) just after the impact & `v_(2)=` velocity of (M+M+m) just of the maximum compression of the SPRING. `RARR v_(1)=(mv_(0))/(M+m) & V_(2)=(mv_(0))/(2M+m)` The loss of K.E. between the two portion (a) and (b) is given as, `Delta KE=(1)/(2)(M+m)v_(1)^(2)-(1)/(2)(2M+m)v_(2)^(2)`  `rArr Delta KE=(m^(2)v_(0)^(2))/(2)[(1)/((M+m))-(1)/((2M+m))]=(m^(2)Mv_(0)^(2))/(2(M+m)(2M+m))` From conservation of mechanical energy `rArr (1)/(2)kx^(2)=(m^(2)Mv_(0)^(2))/(2(M+m)(2M+m))` `rArr x = SQRT((m^(2)Mv_(0)^(2))/(2k(M+m)(2M+m)))=mv_(0)sqrt((M)/(2k(M+m)(2M+m)))` |
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